Question: The following equation is true for all real values of $k$ for which the expression on the left is defined, and $A$ is a polynomial expression. $\dfrac{6k^2-36k}{A}\cdot\dfrac{k^2+12k+36}{k^2-36}=1$ What is $A$ ? $A=$
Solution: The left side of the equation is a product of two rational expressions, while the right side is simply $1$. This means that the numerator and the denominator of the resulting product on the left side should cancel out completely. In order to solve for $A$, let's multiply the expressions and simplify as much as we can. We start with factoring the numerators and the denominators of the two expressions. [Why are we doing this?] The numerator, $6k^2-36k$, of the first expression can be factored as $6k(k-6)$ by factoring out $6k$. The numerator, $k^2+12k+36$, of the second expression can be factored as $(k+6)(k+6)$ using the perfect square pattern. The denominator, $k^2-36$, of the second expression can be factored as $(k+6)(k-6)$ using the difference of squares pattern. Now the product looks as follows: $\dfrac{6k(k-6)}{A}\cdot\dfrac{(k+6)(k+6)}{(k+6)(k-6)}$ To find the product of two rational expressions, we multiply across, then simplify: [What's that?] $\begin{aligned} &\phantom{=}\dfrac{6k(k-6)}{A}\cdot\dfrac{(k+6)(k+6)}{(k+6)(k-6)} \\\\\\ &= \dfrac{6k(k-6) \cdot (k+6)(k+6)}{A \cdot (k+6)(k-6)} &\text{Multiply across.}\\\\\\ &= \dfrac{6k{\cancel{(k-6)}} {\cancel{(k+6)}}(k+6)}{A {\cancel{(k+6)}}{\cancel{(k-6)}}} &\text{Cancel out common factors.}\\\\\\\\ &=\dfrac{6k(k+6)}{A} \end{aligned}$ After this simplification, our equation now looks like: $\dfrac{6k(k+6)}{A}=1$ We can conclude that the numerator and the denominator of the expression on the left must be equal. [Why?] In other words, $A=6k(k+6)$, which is equivalent to $6k^2+36k$.